Question #d3c9f

1 Answer
Apr 10, 2015

32.1" m/s "[82^o" below horizontal"]

First divide the initial velocity vector into horizontal components and vertical components.

The horizontal component will be

v_x=vcostheta = 7.0cos53 m/s

I am assuming we are neglecting air resistance so this value will remain constant. v_(fx) = v_(ix) = v_x =4.21 m/s

The vertical component will be

v_y=vsintheta = 7.0sin53 m/s [up]

v_(fy) the final velocity in the y-direction can be determined by

(v_(fy))^2=(v_(iy))^2 + 2gDeltay . . . . substitution gives

v_(fy)=sqrt((7.0sin53)^2 + 2(-9.81)(-50))=31.82 m/s

Now combine the two components to get the final velocity
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v_f=sqrt(v_(fx)^2 + v_(fy)^2) = 32.1 m/s

theta = tan^(-1)((v_"fy")/(v_"fx"))=tan^(-1)((31.82)/(4.21))=82^o